# Calcium chloride dissolves in water to form calcium and chloride ions, producing 83kJ of heat per mole. Calculate final temp. when 0.02 moles of calcium chloride is dissolved into 150ml of water at a temperature initially of 19 degrees?

Jul 5, 2017

${T}_{\text{final" = 21.6^@"C}}$

#### Explanation:

Given:

The process of dissolving $C a C {l}_{2}$ in releases $\left(8.3 \times {10}^{4} \text{J")/("mole of } C a C {l}_{2}\right)$

Compute the heat energy released for $0.02 \text{mole of } C a C {l}_{2}$:

(8.3xx10^4"J")/(cancel("mole of "CaCl_2))(0.02cancel("mole of "CaCl_2))/1 = 1660"J"

This heat energy must be absorbed by the water. The reference Specific Heat gives us a useful equation:

$Q = c m \Delta T$

Substitute $1660 \text{J}$ for Q:

$1660 \text{J} = c m \Delta T$

The reference, also, gives us a value for c:

1660"J" = (4.186"J")/("g "^@"C")mDeltaT

Substitute 150g for m:

1660"J" = (4.186"J")/("g "^@"C")(150"g")DeltaT

Substitute ${T}_{\text{final" - 19^@"C}}$ for $\Delta T$:

1660"J" = (4.186"J")/("g "^@"C")(150"g")(T_"final" - 19^@"C")

Solve for ${T}_{\text{final}}$

${T}_{\text{final" = (1660"J")/((4.186"J")/("g "^@"C")(150"g")) + 19^@"C}}$

${T}_{\text{final" = 21.6^@"C}}$